You are probably right but the proper equilibrium assumption that “more costly trembles are made with significantly smaller probability than less costly ones” is a huge one.
That assumption isn’t really relevant here, since there aren’t actually any second level trembles with probability ϵ^2 in the solution. (Maybe trembling hand perfect equilibrium already gives us this unique solution and I didn’t really need to invoke proper equilibrium, but the latter seems easier to work with.) Instead of talking about technicalities though, let’s just think about this intuitively.
Suppose there is some definite probability of physically making a mistake when Player One intends to choose A. Let’s say when he intends to choose A, there’s probability 1⁄100 for accidentally pressing B and 1⁄100 for accidentally pressing C, and this is common knowledge. Now we can just solve this modified game using standard Nash equilibrium, and the only solution is that Player One has to make his choices so that after taking both “trembling” and deliberate choice into account, the probabilities that Player Two face is that C is twice as likely as B. (I.e., Player One has to deliberately choose C with probability about 1⁄100.) That’s the only way that Player Two would be willing to choose a mixed strategy, which must be the case in order to have an equilibrium.
Hmm, I just realized that Player Two’s strategy in the Nash equilibrium of this modified game is different than in the proper equilibrium of the original game. Because here Player Two has to make his choice so that Player One is indifferent between A and C, whereas in the proper equilibrium Player Two has to make his choices so that Player One is indifferent between B and C.
I think my “intuitive analysis” does make sense, so I’m going to change my mind and say that perhaps proper equilibrium isn’t the right solution concept here...
You are probably right but the proper equilibrium assumption that “more costly trembles are made with significantly smaller probability than less costly ones” is a huge one.
That assumption isn’t really relevant here, since there aren’t actually any second level trembles with probability ϵ^2 in the solution. (Maybe trembling hand perfect equilibrium already gives us this unique solution and I didn’t really need to invoke proper equilibrium, but the latter seems easier to work with.) Instead of talking about technicalities though, let’s just think about this intuitively.
Suppose there is some definite probability of physically making a mistake when Player One intends to choose A. Let’s say when he intends to choose A, there’s probability 1⁄100 for accidentally pressing B and 1⁄100 for accidentally pressing C, and this is common knowledge. Now we can just solve this modified game using standard Nash equilibrium, and the only solution is that Player One has to make his choices so that after taking both “trembling” and deliberate choice into account, the probabilities that Player Two face is that C is twice as likely as B. (I.e., Player One has to deliberately choose C with probability about 1⁄100.) That’s the only way that Player Two would be willing to choose a mixed strategy, which must be the case in order to have an equilibrium.
Yes, I think you are right. I wonder if there is a way of changing the payoffs so there isn’t any trembling mixed equilibrium.
Hmm, I just realized that Player Two’s strategy in the Nash equilibrium of this modified game is different than in the proper equilibrium of the original game. Because here Player Two has to make his choice so that Player One is indifferent between A and C, whereas in the proper equilibrium Player Two has to make his choices so that Player One is indifferent between B and C.
I think my “intuitive analysis” does make sense, so I’m going to change my mind and say that perhaps proper equilibrium isn’t the right solution concept here...